12 Power Amplifiers

12.0 Power Amplifiers 1 of 37




12.0 Power Amplifiers 5 of 37 Class A Power Calculations Pi(dc) = Vcc*Ico Po(ac) = Icrms

2*Rc = (Icp/2 ½ )2*Rc = (Vcp/2 ½ )2/Rc = (Vcpp/2*2 ½ )2/Rc = (Vcpp)2/8Rc = (Icpp)2*Rc /8 and since, Vcpp = Icpp *Rc then, Po(ac) = Vcpp* Icpp /8

12.0 Power Amplifiers 6 of 37 Maximum Power Output and Efficiency If the max output voltage swing is, Vcppmax = Vcc and then, Pomax(ac) = (Vcc)2/8Rc In order for this to be true the bias situation is that VCQ = Vcc/2 and therefore, ICQ = Vcc/2Rc Power input to the circuit is then, Pimax(dc) = Vcc*Vcc/2Rc The efficiency is then, % = Pomax(ac)/Pimax(dc) = 25% But, if the bias is not in the center, the power output is as calculated as Po(ac) = Vcpp* Icpp /8 and Pi(dc) = Vcc*Ico

12.0 Power Amplifiers 7 of 37 The Class A amplifier has a load line and Q-point as shown. Determine the power situation. a) What is the Pi? b) What is the maximum Po when the amp is operated at this Q-point? Pi = VCC*ICQ = 20V * 2mA = 40mW Po = Vrms2/R R = 20/8mA = 2500 = (5/20.5)2/2500 = = 5mW or, Po = (Vcemax – Vcemin)(Icmax – Icmin) /8 = (20-10)(4 - )/8 = 5mW c) What is the maximum Po when the amp is operated at a Q-point of 10v and 4ma? Po = (20 – 0)(8 – 0)/8 = 20mW and Pi = 20V*4mA = 80mW

Ic

Vce 20v 15v

8ma

Qpt

0

0 0

2ma

12.0 Power Amplifiers 8 of 37 For the circuit below determine,

C1

1

Q1

Q2N2222

R1

1k

10.70mA

V220Vdc

R2

350k

55.13uA

VAMPL = 4m

0 a) Pi(dc) , ignore the base current Pi = 20V*10.7mA = 0.214W b) Po(ac) Po = (10.6 – 7.8)2 /1k = 0.952mW ( 2*21/2 )2

12.0 Power Amplifiers 9 of 37 The collector-emitter voltage swing is,

Time

0s 2ms 4ms 6ms 8ms 10msV(R1:1)

7V

8V

9V

10V

11V





The signal output swing is nearly 2x the supply voltage. This is the efficiency improvement.

12.0 Power Amplifiers 14 of 37 Class A Transformer Power Calculations VCEpp = VCEmax – VCEmin ICpp = ICmax – Icmin Po(ac) = (VCEmax – VCEmin)( ICmax – Icmin)/8 Efficiency of amplifier Pi(dc) = Vcc*ICQ % efficiency = 50*((VCEmax – VCEmin) /(VCEmax + VCEmin))2

If VCEmin = 0, then efficiency is max at 50%



12.0 Power Amplifiers 17 of 37 IB = 6mA & Ib p-p = 8mA, swing of +- 4mA for N1//N2 = 3 & RL = 8 RL’ = 72 ohms Then the slope of 1/72 gives an intersection of current axis at, Ic = 140mA + 10V/72 = 140mA + 139mA = 279mA Looking at the graph & setting the max Ib = 6mA + 4mA = 10mA........results in a max of 255mA by eyeballing the plot. The minimum current is when Ib = 6mA – 4mA = 2mA and is 25mA again from estimating the plot. from the plot the min and max Vce’s are found to be 1.7V and 18.3V. Notice that the signal swing is nearly 2X the supply voltage.....this where the efficiency gain occurs. So, the power output is, Po(ac) = (18.3 – 1.7)(255mA – 25mA)/8 = 0.477W Pi(dc) = Vcc * Icq = 10V*140mA = 1.4W & the efficiency is 0.477/1.4 = 34.1% The max for a class A transformer coupled amp is 50%.

12.0 Power Amplifiers 18 of 37 The resistance load of a transistor amplifier is plotted on the transistor characteristics. Determine, a) Input power, P(dc) b) Maximum output power

Q-point

P(dc) = (160mA)(25) = 4 W R = 25/0.4 = 62.5 ohms P(ac) = Vp2/2R = (25-15)2/2*62.5 = 0.8 W

12.0 Power Amplifiers 19 of 37 The load reflected back to the transistor is 50 ohms. Determine the following, a) Load line b) Minimum and maximum output voltages c) Maximum power efficiency for this bias condition

Q-point


Q-point

eff = 50%(VCEmax – VCEmin)( VCEmax + VCEmin)/8 = 50%[(17.5 – 2.5)/(17.5 + 2.5)]2 = 28.125%

12.0 Power Amplifiers 21 of 37 For the transformer-coupled Class A amplifier below the dc base current is 6ma and the signal base current swings from 8ma to 4ma. The turns ratio is 5:1 and the load resistance is 4 ohms. a) Find and plot the Q-point on the collector characteristics below. 12V & 180mA 5pt b) What is the input power including both the base and collector currents? (180+6)*10 = 1860mW = 1.86W 5pt c) What is the power delivered to the load? PL =~ 1/8 *(18 – 7)*(240mA – 125mA) = 0.18W 5pt Loadline 100 5pt 12V

5:1

4 ohms

0

0

200ma

0

400ma

300ma

100ma

5 10 15 20 Vce

2ma

4ma

6ma

8ma

10ma

12ma

0

Ic

12.0 Power Amplifiers 22 of 37 The Class A transformer-coupled power amplifier circuit schematic and simulation results are given below. The schematic with DC collector current is,

TX1

V220Vdc

R1

4

Q1

Q2N2222

55.58mAC1

1

0

0

R2

75k

VAMPL = 0.004

Q1 collector voltage waveform is,

Time

18.0us 18.5us 19.0us 19.5us 20.0usV(TX1:2)

18V

19V

20V

21V

22V

12.0 Power Amplifiers 23 of 37 The collector current waveform is,

Time

18.0us 18.5us 19.0us 19.5us 20.0usI(TX1:1)

48mA

52mA

56mA

60mA

64mA

12.0 Power Amplifiers 24 of 37 4 ohm resistor voltage waveform is,

Time

18.0us 18.5us 19.0us 19.5us 20.0usV(TX1:3)

-200mV

-100mV

0V

100mV

200mV

Determine the following, a) Pi(dc) ignore the base current Pi = 20V*55.58mA = W

b) Po(ac) using the collector voltage & current waveform values. Po = (1/8)(21.5V – 18.4V)*(61.8mA – 49.8mA) = W c) Po(ac) using the load voltage waveform values (show calculations). Po = [(200mV + 185mV)/2*2 ½ ]2/4

12.0 Power Amplifiers 25 of 37 Assume the power amplifiers in the questions below are operating at a Q-point such that maximum power output could be obtained if desired. Determine for each power amplifier circuit the following, a) Power Input b) Power Output Show your work; i.e., the formulas/algebra you used for each problem. 1. A resistor-load power amplifier has a voltage output swing peak voltage equal to 0.4Vcc. Pi = Vcc * Vcc/2R = Vcc

2/2R

Po = (0.4Vcc/(21/2

))2/R

2. A 1:1 transformer-load power amplifier has a voltage output swing peak voltage equal to 0.8Vcc. Pi = Vcc*Vcc/R = Vcc

2/R

Po = (0.8Vcc/21/2

)2/R

3. A Class B (push-pull) power amplifier has a voltage output swing peak voltage equal to 0.7Vcc. Pi = Vcc*2(0.7)Vcc/πR Po = (0.7Vcc/2

1/2)

2/R

12.0 Power Amplifiers 26 of 37 Class B amplifier - eliminate the bias current.



12.0 Power Amplifiers 29 of 37 Class B Power Calculations Input Power = Pi(dc) = Vcc*Idc For each half cycle current goes through the power supply, Vcc. For a half wave rectifier the average current is, Idc = Ip/π and since this current flows on each half cycle the average current is Idc = 2Ip/π = 2* VLp/RL π Pi(dc) = 2*Vcc* VLp/RL π Output Power Po(ac) = VLrms

2/RL = VLp

2/2RL = VLpp

2/8RL

12.0 Power Amplifiers 30 of 37 Efficiency % = 100%*Po(ac)/Pi(dc) = 100%*( VLp

2/2RL)/( 2* VLp/RL π) 100%*π*(VLp/4Vcc) and if VLp = Vcc/2 then % = π/4 = 78.5% Transistor Power P2Q = Pi(dc) - Po(ac) PQ = ½ *P2Q


12.0 Power Amplifiers 32 of 37 Power Amplifiers Resistor Load Pi = Vcc * Vcc/2R = Vcc2/2R Po = (Vcc/(2*21/2 ))2/R = Vcc2/8R %efficiency max = 25% Transformer Load 1:1 turns ratio Pi = Vcc*Vcc/R = Vcc2/R Po = (Vcc/21/2 )2/R = Vcc2/2R %efficiency max = 50% Push-Pull Pi = 2Vcc*Vcc/πR Po = (Vcc/21/2)2/R = Vcc2/2R %efficiency max = (1/2)/(2/ π) = 25π %


Summary of Power Amplifiers Draw 3 circuits Output power - voltage and current Signal swing (peak-to-peak / 2)/sqrt of 2 for rms Ipp*Vpp/8 Input power Average Current*Average Voltage Supply

Vpp Ipp OutputPower

Avg Current

Voltage Supply

Input Power

%efficiency

Class A Vcc Vcc/R Vcc2/8R Vcc/2R Vcc Vcc2/2R 25Class A Transformer

2Vcc 2Vcc/R’ 4Vcc2/8R’ Vcc/R Vcc Vcc2/R’ 50

Class B 2Vcc 2Vcc/R 4Vcc2/8R 2Vcc/πR Vcc 2Vcc2/πR 100π/4

12.0 Power Amplifiers 34 of 37 Amplifier Distortion Harmonic Distortion % nth harmonic distortion

= % Dn where n = 2, 3, 4,…….. = 100%* |An|/|A1|

% THD = square root of sum of squares of Dn Power of Signal Having Distortiion P1 = I12*Rc/2 P = (I12+ I22+ I22 + …..)Rc/2 = (1 + THD2)P1




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12 Power Amplifiers